# Plane curves of bounded curvature

Here is a "conjecture" that should be known (but I have not found any good reference to it): consider the class $\cal F$ of all $C^2([0,1])$ functions $f$ with the property $f(0)=f'(0)=f''(0)=0$ and $f(1)-1=f'(1)=f''(1)=0$. We can think of the graph of $y=f(x)$ as a transfer path between two parallel and horizontal railroads, one arriving at (0,0) and one starting at (1,1).

Such a function in $\cal F$ is the polynomial $P(x)=6x^5-15x^4+10x^3$ and the (absolute value) maximum curvature of $P$ (the general the formula for the curvature is $k_f(x)=\frac{f''(x)}{(1+f'(x)^2)^{\frac{3}{2}}}$) is a little over 4. There are other simple functions in $\cal F$ with lower maximum curvature (and it is perhaps a good exercise in Calculus III courses, to try find others, since there are infinite families in $\cal F$ that one can come up with).

For example, the piece-wise polynomial,

$$Q(x)=\begin{cases} 8x^3(1-x), \ \ \ x\in [0,\frac{1}{2}],\\ 1-8(1-x)^3x, \ \ \ \ x\in [\frac{1}{2},1],\end{cases}$$ is in $\cal F$ and it has a maximum curvature which is below $4$.

Is it true that $$m:=\underset{f\in \cal F}{inf}\left( \underset{f\in [0,1]}{max}{|k_f(x)}|\right)=2 ?$$

One can use two quarter circles of radius 1/2 and construct a path connecting the two points. The path is not quite what is required but perhaps it can be "fixed" and that will give m≤2. The other part seems to be more difficult to prove. In other words, if the curvature is less than 2, then one cannot connect those points with a curve y=f(x). If one drops this requirement and asks only for a curve in the plane, I assume the answer is m=0.

The following function, which appears from the calculation of the (normalized) area of the segment of a circle of radius $\frac{1}{2}$ in terms of its height $x$ (there is a small change of a constant in the middle though), $$f(x)=[ \arccos(1-2x)+(1-2x)(x-x^2)^{\frac{1}{2}}]/\pi,\ x\in [0,1]$$ has an inverse. Taking $g=f^{-1}$, one can check that $g\in \cal F$ and its maximum curvature which is below 2.33

• What does lead you to think about that estimate?
– Edu
Mar 16, 2018 at 14:59
• I have edited the initial post to reflect this question. Mar 18, 2018 at 23:55
• Apply the Mean Value Theorem twice
– Del
Mar 19, 2018 at 11:03
• I could only get that $m>2-\sqrt{2}$ with Mean Value applied several times .... Mar 26, 2018 at 19:48

We will show $m \ge 2$ and argue the actual value of $m$ is $2$.

View the graph of $f$ as a path in $\mathbb{R}^2$:

$$\gamma:\quad [0,1] \ni t\quad\mapsto\quad (x,y) = (t,f(t)) \in \mathbb{R}^2$$

Parameterize $\gamma$ by its arc-length measured from $(0,0)$, $$s(x) = \int_0^x \sqrt{1+f'(t)^2} dt$$

Let $\theta$ be the angle between tangent vector of $\gamma$ and $x$-axis, $$(\cos\theta,\sin\theta) = \left(\frac{dx}{ds}, \frac{dy}{ds}\right) = \left(\frac{1}{\sqrt{1+f'^2}},\frac{f'}{\sqrt{1+f'^2}}\right)$$

Let $K$ be the maximum of absolute value of $\frac{d\theta}{ds}$. i.e.

$$K = K(f) \stackrel{def}{=} \max_{t\in [0,1]} \left|\frac{d\theta}{ds}\right| = \max_{t\in [0,1]} \left|\frac{f''(t)}{(1+f'(t)^2)^{3/2}}\right|$$

Let us consider what will happen when $K < 2$.

Since $f'(0) = 0 \implies \theta(0) = 0$,

$$\left|\frac{d\theta}{ds}\right| \le K \implies |\theta(s)| = \left|\int_0^s \frac{d\theta}{ds} ds\right| \le \int_0^s K ds = Ks$$

As long as $s \le \frac{\pi}{2K}$, we will have

$$\cos\theta(s) \ge \cos(Ks)\quad\text{ and }\quad |\sin\theta(s)| \le \sin(Ks)$$

Since $x(0) = y(0) = 0$, this leads to \begin{align} x(s) &= \int_0^s \cos\theta(s) ds \ge \int_0^s \cos(K\tau)d\tau = \frac{\sin(Ks)}{K}\tag{*1a}\\ |y(s)| &= \left|\int_0^s \sin\theta(s) ds\right| \le \int_0^s \sin(K\tau) d\tau = \frac{1-\cos(Ks)}{K}\tag{*1b} \end{align}

When $K < 2$, we have $\frac1K > \frac12$. $(*1a)$ tells us $x(s)$ will reach $\frac12$ at some $s = s_* < \frac{\pi}{2K}$. Together with $(*1b)$, we find $$f\left(\frac12\right) = y(s_*) \le \frac{1-\cos(Ks_*)}{K} = \frac{1 - \sqrt{1 - \sin(Ks_*)^2}}{K} \le \frac{1 - \sqrt{1 - (Kx(s_*))^2}}{K}$$ Notice $(K + \sqrt{4-K^2})^2 = 4 + 2K\sqrt{4-K^2} > 4 \implies K + \sqrt{4-K^2} > 2$, we find

$$f\left(\frac12\right) \le \frac{1 - \sqrt{1 - (Kx(s_*))^2}}{K} = \frac{2 - \sqrt{4 - K^2}}{2K} < \frac12$$

Instead of parameterize $\gamma$ using arc-length measured from $(0,0)$. we can parameterize $\gamma$ using arc-length measured from $(1,1)$. Using essentially the same argument as above but on the portion of $\gamma$ at $x \ge \frac12$, we can show that when $K < 2$, $f(\frac12) > \frac12$. From this, we can conclude it is impossible for $|\frac{d\theta}{ds}| < 2$ over the whole path. In other words $$\bbox[border:1px solid blue;padding:1em]{m = \inf_{f \in \mathcal{F}} K(f) \ge 2}$$

About the actual value of $m$, it should be $2$.

Consider following curve consisting of two quarter circular arcs:

• one centered at $(0,\frac12)$ with radius $\frac12$ joining $(0,0)$ to $(\frac12,\frac12)$.

• another centered at $(1,\frac12)$ with radius $\frac12$ joining $(\frac12,\frac12)$ to $(1,1)$.

This curve close to give a $f$ that we want. Aside from the point $(\frac12,\frac12)$, we have $\left|\frac{d\theta}{ds}\right| = 2$. This curve do have some minor problems. First, $f''(0), f''(1) \ne 0$. Second, $f'$ diverges at $\frac12$ and hence $f$ fails to belong to $C^2([0,1])$.

It is easy to modify $f$ to force $f''(0)$ and $f''(1)$ to vanish. If one increase the bound for $\left|\frac{d\theta}{ds}\right|$ near $x=0$ and $x=1$ a little bit, one will have enough freedom to smooth out the singularity of $f'$ at $(\frac12,\frac12)$. This suggests one can construct a $f \in C^2([0,1])$ with $K(f)$ as close to $2$ as one desire.

• Very nice! Thanks for this clever solution. Now, it begs the question of what happens if we change $\cal F$ to $${\cal F}_t:=\{f\in C^2[0,1]| f(0)=f'(0)=0 and f(1)=1, f'(1)=t,f''(1)=0\}, \ t\in \mathbb R$$ Mar 30, 2018 at 19:13
• @EugenIonascu given any bound $K$ of the curvature, the portion of path $\gamma$ near $(0,0)$ is avoiding two circles of radius $\frac{1}{K}$ touching $(0,0)$ (with tangent in $x$-axis). Similar thing happens for the portion near $(1,1)$, it will avoiding two circles of radius $\frac{1}{K}$ touching $(1,1)$ (with tangent in the direction $(1,t)$). For $t$ not too large, the problem should reduce to a geometry problem of finding the smallest $K$ where the two admissible region near the ends connect to each other. Mar 30, 2018 at 19:28
• I agee! What I got is $$k_f=\frac{1}{2}\left( \cos(u)-\sin(u)+1+\sqrt{8-2(1+\sin(u))(1+\cos(u))} \right),$$ where $t=\tan u$, at least if $u\in [0,\frac{\pi}{2}]$ Mar 30, 2018 at 20:41
• @EugenIonascu I reproduced your expression of $k_f$ and the curve looks good when implemented on GeoGebra. Mar 30, 2018 at 22:35
• @EugenIonascu For $u \in (-\frac{\pi}{2},0)$, the curve constructed by joining the two circle arcs (with curvature given by your expression) is moving backward in $x$. Since $\gamma$ is supposed to be a graph of $f$, this cannot happen, the radius of the circles need to be smaller. The limiting condition becomes the two circles shares a common vertical tangent. $k_f$ should be $2 + \sin|u|$ when $u \in (-\frac{\pi}{2},0)$. Mar 30, 2018 at 22:57